Source: Rings With a Polynomial Identity, Irving Kaplansky
The Lemma:
Suppose that $\mathbb{A}$ is an $\mathbb{F}$-algebra, where $\mathbb{F}$ is a field. Then, suppose that $\mathbb{A}$ satisfies a polynomial identity $f=0$ so that the degree of each variable $x_i$ of $g$ is no more than $n$. Then, if $\mid \mathbb{F} \mid \geq n+1$, then $\mathbb{A}$ satisfies a polynomial identity which is homogenous in each of its variables.
I am having trouble understanding part of the proof.
The proof:
Write$ f= \sum f_i$ where the degree of $x_1$ in $f_i$ is $i$. Then, replace $x_1$ by $\lambda_1 x_1$, where $\lambda_1$ is from the field $\mathbb{F}$. We then immediatly see that $\sum \lambda_1^i f_i=0$. We do this $n$ times for $n$ distinct nonzero scalars $\lambda_i \in \mathbb{F}$.
Taking the Vandermonde determinant with the $\lambda_i$ above, we see that $f_i=0$. This part is fine, because the equations obtained above are just the product of the Vandermonde matrix with the column of the $f_i$, and since the $\lambda_i$ are distinct, the determinant is nonzero, so the matrix is invertible.
By repeating this process for each of the variables, we obtain an identity that is homogenous in each variable.
The only part that I'm not understanding is why the resulting identity is homogenous.
Thanks, -Chris